Question

1. determine intervals of increasing, decreasing, or constant. find turning points (relative maxima or minima). given:

a) ( f(x)=2(x - 5)^{2}(x + 7) )

Explanation:

Step1: Expand the function

$$\begin{align*} f(x)&=2(x - 5)^{2}(x + 7)\\ &=2(x^{2}-10x + 25)(x + 7)\\ &=2(x^{3}+7x^{2}-10x^{2}-70x + 25x+175)\\ &=2(x^{3}-3x^{2}-45x + 175)\\ &=2x^{3}-6x^{2}-90x + 350 \end{align*}$$

Step2: Find the derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=6x^{2}-12x - 90\). Factor out \(6\): \(f^\prime(x)=6(x^{2}-2x - 15)\). Then factor the quadratic: \(f^\prime(x)=6(x - 5)(x+ 3)\)

Step3: Find critical points

Set \(f^\prime(x)=0\), so \(6(x - 5)(x + 3)=0\). Solving for \(x\), we get \(x=-3\) and \(x = 5\)

Step4: Test intervals

• For \(x<-3\), let \(x=-4\). Then \(f^\prime(-4)=6(-4 - 5)(-4 + 3)=6\times(-9)\times(-1)=54>0\). So \(f(x)\) is increasing on \((-\infty,-3)\)
• For \(-3
• For \(x>5\), let \(x=6\). Then \(f^\prime(6)=6(6 - 5)(6 + 3)=6\times1\times9 = 54>0\). So \(f(x)\) is increasing on \((5,\infty)\)

Step5: Find turning points

• When \(x=-3\), \(f(-3)=2(-3 - 5)^{2}(-3 + 7)=2\times64\times4 = 512\). So the relative maximum is at the point \((-3,512)\)
• When \(x = 5\), \(f(5)=2(5 - 5)^{2}(5 + 7)=0\). So the relative minimum is at the point \((5,0)\)

Answer:

• Increasing intervals: \((-\infty,-3)\cup(5,\infty)\)
• Decreasing interval: \((-3,5)\)
• Relative maximum: \((-3,512)\)
• Relative minimum: \((5,0)\)